Calculate the pH: a) A solution containing 0.0282 M maleic acid and 0.040 M diso

Question

Calculate the pH:

a) A solution containing 0.0282 M maleic acid and 0.040 M disodium maleate. The Ka values for maleic acid are 1.20 × 10-2 (Ka1) and 5.37 × 10-7 (Ka2).

b) A solution containing 0.0307 M phthalic acid and 0.018 M potassium hydrogen phthalate. The Ka values for phthalic acid are 1.12 × 10-3 (Ka1) and 3.90 × 10-6 (Ka2).

Calculate the pH of the following solutions. Mapoob a) A solution containing 0.0282 M maleic acid and 0.040 M disodium maleate. The Ka values for maleic acid are 1.20 × 102 (Kal) and 5.37 × 10' (Ke). Number PH- b) A solution containing 0.0307 M phthalic acid and 0.018 M potassium hydrogen phthalate. The Ka values for phthalic acid are 1.12 × 10-3 (Kat) and 3.90 x 10° (Ko). Number pH- ID

Explanation / Answer

a) A solution containing 0.0282 M maleic acid and 0.040 M disodium maleate. The Ka values for maleic acid are 1.20 × 10-2 (Ka1) and 5.37 × 10-7 (Ka2).

First, identify acidic mmoles:

M of acid = n*M = 0.0282*2 = 0.0564

Now, M of base, Maleate:

M of base = 0.04 M = 0.04

clearly, there is not enough base to neutralize the acid

M = 0.04- 0.0282 = 0.0118 M of maleate is left, after neutralizing the first acid

M of 2nd acid = 0.0282

pH= pKa2 + log(maleate/ acid)

pH = -log(5.37*10^-7) + log(0.0118/0.0282)

pH = 5.8916

b) A solution containing 0.0307 M phthalic acid and 0.018 M potassium hydrogen phthalate. The Ka values for phthalic acid are 1.12 × 10-3 (Ka1) and 3.90 × 10-6 (Ka2).

similarly:

pH = pKa1 + log(conjguate/acid)

pH = -log(1.12*10^-3) + log(0.018 / 0.0307)

pH = 2.718

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